3.275 \(\int \frac{a+b \log (c (d+e x)^n)}{\sqrt{2+g x^2}} \, dx\)

Optimal. Leaf size=326 \[ -\frac{b n \text{PolyLog}\left (2,-\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{d^2 g+2 e^2}}\right )}{\sqrt{g}}-\frac{b n \text{PolyLog}\left (2,-\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{\sqrt{d^2 g+2 e^2}+d \sqrt{g}}\right )}{\sqrt{g}}+\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{d^2 g+2 e^2}}+1\right )}{\sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{\sqrt{d^2 g+2 e^2}+d \sqrt{g}}+1\right )}{\sqrt{g}}+\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )^2}{2 \sqrt{g}} \]

[Out]

(b*n*ArcSinh[(Sqrt[g]*x)/Sqrt[2]]^2)/(2*Sqrt[g]) - (b*n*ArcSinh[(Sqrt[g]*x)/Sqrt[2]]*Log[1 + (Sqrt[2]*e*E^ArcS
inh[(Sqrt[g]*x)/Sqrt[2]])/(d*Sqrt[g] - Sqrt[2*e^2 + d^2*g])])/Sqrt[g] - (b*n*ArcSinh[(Sqrt[g]*x)/Sqrt[2]]*Log[
1 + (Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sqrt[2]])/(d*Sqrt[g] + Sqrt[2*e^2 + d^2*g])])/Sqrt[g] + (ArcSinh[(Sqrt[g]
*x)/Sqrt[2]]*(a + b*Log[c*(d + e*x)^n]))/Sqrt[g] - (b*n*PolyLog[2, -((Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sqrt[2]]
)/(d*Sqrt[g] - Sqrt[2*e^2 + d^2*g]))])/Sqrt[g] - (b*n*PolyLog[2, -((Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sqrt[2]])/
(d*Sqrt[g] + Sqrt[2*e^2 + d^2*g]))])/Sqrt[g]

________________________________________________________________________________________

Rubi [A]  time = 0.420563, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {215, 2404, 12, 5799, 5561, 2190, 2279, 2391} \[ -\frac{b n \text{PolyLog}\left (2,-\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{d^2 g+2 e^2}}\right )}{\sqrt{g}}-\frac{b n \text{PolyLog}\left (2,-\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{\sqrt{d^2 g+2 e^2}+d \sqrt{g}}\right )}{\sqrt{g}}+\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{d^2 g+2 e^2}}+1\right )}{\sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{\sqrt{d^2 g+2 e^2}+d \sqrt{g}}+1\right )}{\sqrt{g}}+\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )^2}{2 \sqrt{g}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x)^n])/Sqrt[2 + g*x^2],x]

[Out]

(b*n*ArcSinh[(Sqrt[g]*x)/Sqrt[2]]^2)/(2*Sqrt[g]) - (b*n*ArcSinh[(Sqrt[g]*x)/Sqrt[2]]*Log[1 + (Sqrt[2]*e*E^ArcS
inh[(Sqrt[g]*x)/Sqrt[2]])/(d*Sqrt[g] - Sqrt[2*e^2 + d^2*g])])/Sqrt[g] - (b*n*ArcSinh[(Sqrt[g]*x)/Sqrt[2]]*Log[
1 + (Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sqrt[2]])/(d*Sqrt[g] + Sqrt[2*e^2 + d^2*g])])/Sqrt[g] + (ArcSinh[(Sqrt[g]
*x)/Sqrt[2]]*(a + b*Log[c*(d + e*x)^n]))/Sqrt[g] - (b*n*PolyLog[2, -((Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sqrt[2]]
)/(d*Sqrt[g] - Sqrt[2*e^2 + d^2*g]))])/Sqrt[g] - (b*n*PolyLog[2, -((Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sqrt[2]])/
(d*Sqrt[g] + Sqrt[2*e^2 + d^2*g]))])/Sqrt[g]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 2404

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/Sqrt[(f_) + (g_.)*(x_)^2], x_Symbol] :> With[{u = Int
Hide[1/Sqrt[f + g*x^2], x]}, Simp[u*(a + b*Log[c*(d + e*x)^n]), x] - Dist[b*e*n, Int[SimplifyIntegrand[u/(d +
e*x), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && GtQ[f, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 5799

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Subst[Int[((a + b*x)^n*Cosh[x
])/(c*d + e*Sinh[x]), x], x, ArcSinh[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c (d+e x)^n\right )}{\sqrt{2+g x^2}} \, dx &=\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}-(b e n) \int \frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}{\sqrt{g} (d+e x)} \, dx\\ &=\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}-\frac{(b e n) \int \frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}{d+e x} \, dx}{\sqrt{g}}\\ &=\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}-\frac{(b e n) \operatorname{Subst}\left (\int \frac{x \cosh (x)}{\frac{d \sqrt{g}}{\sqrt{2}}+e \sinh (x)} \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )\right )}{\sqrt{g}}\\ &=\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )^2}{2 \sqrt{g}}+\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}-\frac{(b e n) \operatorname{Subst}\left (\int \frac{e^x x}{e e^x+\frac{d \sqrt{g}}{\sqrt{2}}-\frac{\sqrt{2 e^2+d^2 g}}{\sqrt{2}}} \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )\right )}{\sqrt{g}}-\frac{(b e n) \operatorname{Subst}\left (\int \frac{e^x x}{e e^x+\frac{d \sqrt{g}}{\sqrt{2}}+\frac{\sqrt{2 e^2+d^2 g}}{\sqrt{2}}} \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )\right )}{\sqrt{g}}\\ &=\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )^2}{2 \sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (1+\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (1+\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}+\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}+\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}+\frac{(b n) \operatorname{Subst}\left (\int \log \left (1+\frac{e e^x}{\frac{d \sqrt{g}}{\sqrt{2}}-\frac{\sqrt{2 e^2+d^2 g}}{\sqrt{2}}}\right ) \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )\right )}{\sqrt{g}}+\frac{(b n) \operatorname{Subst}\left (\int \log \left (1+\frac{e e^x}{\frac{d \sqrt{g}}{\sqrt{2}}+\frac{\sqrt{2 e^2+d^2 g}}{\sqrt{2}}}\right ) \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )\right )}{\sqrt{g}}\\ &=\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )^2}{2 \sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (1+\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (1+\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}+\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}+\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}+\frac{(b n) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{e x}{\frac{d \sqrt{g}}{\sqrt{2}}-\frac{\sqrt{2 e^2+d^2 g}}{\sqrt{2}}}\right )}{x} \, dx,x,e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}\right )}{\sqrt{g}}+\frac{(b n) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{e x}{\frac{d \sqrt{g}}{\sqrt{2}}+\frac{\sqrt{2 e^2+d^2 g}}{\sqrt{2}}}\right )}{x} \, dx,x,e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}\right )}{\sqrt{g}}\\ &=\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )^2}{2 \sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (1+\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}-\frac{b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \log \left (1+\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}+\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}+\frac{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{\sqrt{g}}-\frac{b n \text{Li}_2\left (-\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}-\frac{b n \text{Li}_2\left (-\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}+\sqrt{2 e^2+d^2 g}}\right )}{\sqrt{g}}\\ \end{align*}

Mathematica [A]  time = 0.231515, size = 275, normalized size = 0.84 \[ \frac{-2 b n \text{PolyLog}\left (2,\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{\sqrt{d^2 g+2 e^2}-d \sqrt{g}}\right )-2 b n \text{PolyLog}\left (2,-\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{\sqrt{d^2 g+2 e^2}+d \sqrt{g}}\right )+\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right ) \left (2 a+2 b \log \left (c (d+e x)^n\right )-2 b n \log \left (\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{d \sqrt{g}-\sqrt{d^2 g+2 e^2}}+1\right )-2 b n \log \left (\frac{\sqrt{2} e e^{\sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )}}{\sqrt{d^2 g+2 e^2}+d \sqrt{g}}+1\right )+b n \sinh ^{-1}\left (\frac{\sqrt{g} x}{\sqrt{2}}\right )\right )}{2 \sqrt{g}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])/Sqrt[2 + g*x^2],x]

[Out]

(ArcSinh[(Sqrt[g]*x)/Sqrt[2]]*(2*a + b*n*ArcSinh[(Sqrt[g]*x)/Sqrt[2]] - 2*b*n*Log[1 + (Sqrt[2]*e*E^ArcSinh[(Sq
rt[g]*x)/Sqrt[2]])/(d*Sqrt[g] - Sqrt[2*e^2 + d^2*g])] - 2*b*n*Log[1 + (Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sqrt[2]
])/(d*Sqrt[g] + Sqrt[2*e^2 + d^2*g])] + 2*b*Log[c*(d + e*x)^n]) - 2*b*n*PolyLog[2, (Sqrt[2]*e*E^ArcSinh[(Sqrt[
g]*x)/Sqrt[2]])/(-(d*Sqrt[g]) + Sqrt[2*e^2 + d^2*g])] - 2*b*n*PolyLog[2, -((Sqrt[2]*e*E^ArcSinh[(Sqrt[g]*x)/Sq
rt[2]])/(d*Sqrt[g] + Sqrt[2*e^2 + d^2*g]))])/(2*Sqrt[g])

________________________________________________________________________________________

Maple [F]  time = 0.852, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c \left ( ex+d \right ) ^{n} \right ) ){\frac{1}{\sqrt{g{x}^{2}+2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))/(g*x^2+2)^(1/2),x)

[Out]

int((a+b*ln(c*(e*x+d)^n))/(g*x^2+2)^(1/2),x)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(g*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{g x^{2} + 2} b \log \left ({\left (e x + d\right )}^{n} c\right ) + \sqrt{g x^{2} + 2} a}{g x^{2} + 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(g*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

integral((sqrt(g*x^2 + 2)*b*log((e*x + d)^n*c) + sqrt(g*x^2 + 2)*a)/(g*x^2 + 2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \log{\left (c \left (d + e x\right )^{n} \right )}}{\sqrt{g x^{2} + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))/(g*x**2+2)**(1/2),x)

[Out]

Integral((a + b*log(c*(d + e*x)**n))/sqrt(g*x**2 + 2), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{\sqrt{g x^{2} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))/(g*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)/sqrt(g*x^2 + 2), x)